Isaac is studying relationships between organisms in the phylum Chordata, which includes humans and all other animals with a backbone. Beginning with domain as 1 and then counting down the levels in the modern system of classification, what level is "phylum”? 2

3

4

5

Phylum is at level 2.

Explanation:

All living organisms are classified, and the reasons for this are so that they can be examined better,to know and discover their relations,their evolutionary paths etc. while in the past there have been some more simplified classifications, the modern one is complex and well established,based on the characteristics of the organisms(living and died out).

when it comes to the domains,there are nine of them,each representing one specific thing,or rather characteristic. Those nine domain(in order) are kingdom,phylum,class,order,suborder,family,subfamily,genus,and species. As an example we can take the snow leopard:

Kingdom:Animalia(an animal)

Phylum:chordata(vertebrate)

Class:Mammalia(mammal)

Order:carnivora(carnivore)

Suborder:Feliformia(cat-like)

Family:Felidae(true cat)

Subfamily:pantherianae(big cat branch)

Genus:panthera(big cat)

Species:P.uncia(snow leopard)

I hope it's helpful!

Phylum is at level 2.

All living organisms are classified, and the reasons for this are so that they can be examined better, to know and discover their relations, their evolutionary paths etc. while in the past there have been some more simplified classifications, the modern one is complex and well established, based on the characteristics of the organisms(living and died out).

Explanation: Kingdom: Animalia(an animal)

Phylum: Chordata (vertebrate)

Class: Mammalia(mammal)

Order: Carnivora(carnivore)

Suborder: Feliformia (also Feloidea) (cat-like)

Family: Felidae(true cat)

Subfamily: pantherine(big cat branch)

Pls give the first person brainiest I'm just doing this for a challenge

Genus:panthera(big cat)

Species:P.uncia(snow leopard)

we solve the problem using the hardy-wineberg equation.

let us start with the homozygous recessive proportion of individuals with sickle cell anaemia,

we have the proportion as q² = 0.25.

∴ q = 0.5 (frequency of hbs in population)

further,

solving for the frequency of normal cell by using the fact that,

p + q = 1

∴ p = 1 - 0.5 = 1 - 0.5 = 0.5

now we can calculate 2pq in p² + 2pq + q², which is the frequency for heterozygous carriers of sickle cell anaemia,

∴ 2pq = 2 x 0.5 x 0.5 = 0.5

and the homozygous normal cell individuals proportion is given by the expression,

∴ p² = 0.5² = 0.25

hence, the carriers for sickle cell anaemia are in proportion, 0.5, or 50%.